Let \(j\in \left\lbrace 1,\ldots ,r\right\rbrace \)  with \(\left\Vert e_j(\tilde{t},\zeta (\tilde{t})) \right\Vert _{ }=\psi _j(\tilde{t})\) .
Recalling the definition of the Lebesgue integral, see e.g. [1]},
\(\int _{I^{T}_{\hat{t}}}\ell (t,\zeta (t),u(t))\,{\rm d} t < \infty \)  implies \(\int _{I^{T}_{\hat{t}}}(\ell (t,\zeta (t),u(t)))^+\,{\rm d} t < \infty \)  where \((\ell (t,\zeta (t),u(t)))^+:=\max \left\lbrace (\ell (t,\zeta (t),u(t))),0\right\rbrace \) .
Note that \(\left\Vert e_i(\tilde{t},\zeta (\tilde{t})) \right\Vert _{ }<\psi _i(\tilde{t})\)  for all \(t\in [\hat{t},\tilde{t})\) 
and for all \(i=1,\ldots , r\) . Therefore,
\(&\int _{\hat{t}}^{\tilde{t}}\tfrac{1}{1- \left\Vert e_j(t,\zeta (t)) \right\Vert _{ }^2/\psi _j(t)^2}{\rm d}{t}\le \int _{\hat{t}}^{\tilde{t}}\sum \limits _{i=1}^r\tfrac{1}{1- \left\Vert e_i(t,\zeta (t)) \right\Vert _{ }^2/\psi _i(t)^2} {\rm d}{t}\!\\&\le \!\!\int _{I^{T}_{\hat{t}}} \!\!\left(\sum \limits _{i=1}^r\tfrac{1}{1- \left\Vert e_j(t,\zeta (t)) \right\Vert _{ }^2/\psi _i(t)^2}\right)^{\!+}\!\! {\rm d}{t}\!\\&\le \!\!\int _{I_{\hat{t}}^T} \!\!\left(\sum \limits _{i=1}^r\tfrac{1}{1- \left\Vert e_i(t,\zeta (t)) \right\Vert _{ }^2/\psi _i(t)^2}-r + \lambda _u\left\Vert u(t) \right\Vert _{ }^2\!\!\right)^{\!+}\!\!{\rm d}{t}+Tr\\&=\!\!\int _{I_{\hat{t}}^T}( \ell (t,\zeta (t),u(t)))^+\!{\rm d}{t}+Tr<\infty .\) 
